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Hint: Atomic radius is defined as the distance from the center of the nucleus to the outermost shell that contains electrons. But, the radius of fluorine is measured in terms of covalent radius and the radius of neon in terms of Van der Waals radius. Also, it is known that Van der Waals radius is greater than the covalent radius. Thus, the radius of fluorine will be smaller than that of neon.
Complete step by step answer:
-Atomic radius can be defined as it is the distance from the center of the nucleus to the point up to which the density of the electron cloud is maximum or as the distance from the center of the nucleus to the outermost shell that contains electrons.
-The Covalent and Van der Waals radii both decrease as the atomic number increases on moving from left to right in a period. The alkali metals which are placed in the extreme left of the periodic table have the largest size in a period. The halogens which are placed in the extreme right of the periodic table have the smallest size. The atomic size of nitrogen is the smallest because, after nitrogen, the atomic size increases for oxygen and then decreases for fluorine. The size of atoms of inert gases is larger than those of the Group 17 halogens.
-The atomic radius unexpectedly increases as we move from halogens to the inert gas elements because inert gases have completely filled orbitals. Also, the atomic size of halogens is expressed in terms of Van der Waals radius since they do not form covalent bonds. And as we know, Van der Waals radius is larger than the covalent radius. Therefore, the atomic size of inert gas in a period is much higher than that of the halogens.
-Thus, from the given options, it can be concluded that the atomic radius of fluorine will be 0.72 (in angstrom unit) and for the neon atom, it will be higher that is 1.60 (in angstrom unit).
Hence, option (A) will be the correct answer.
Note:
We can also look at the fact that fluorine has 5 electrons in its 2p orbital which are 1 less than required for ideal inert gas configuration, which is responsible for its high nuclear charge and smaller size than that of neon.
Complete step by step answer:
-Atomic radius can be defined as it is the distance from the center of the nucleus to the point up to which the density of the electron cloud is maximum or as the distance from the center of the nucleus to the outermost shell that contains electrons.
-The Covalent and Van der Waals radii both decrease as the atomic number increases on moving from left to right in a period. The alkali metals which are placed in the extreme left of the periodic table have the largest size in a period. The halogens which are placed in the extreme right of the periodic table have the smallest size. The atomic size of nitrogen is the smallest because, after nitrogen, the atomic size increases for oxygen and then decreases for fluorine. The size of atoms of inert gases is larger than those of the Group 17 halogens.
-The atomic radius unexpectedly increases as we move from halogens to the inert gas elements because inert gases have completely filled orbitals. Also, the atomic size of halogens is expressed in terms of Van der Waals radius since they do not form covalent bonds. And as we know, Van der Waals radius is larger than the covalent radius. Therefore, the atomic size of inert gas in a period is much higher than that of the halogens.
-Thus, from the given options, it can be concluded that the atomic radius of fluorine will be 0.72 (in angstrom unit) and for the neon atom, it will be higher that is 1.60 (in angstrom unit).
Hence, option (A) will be the correct answer.
Note:
We can also look at the fact that fluorine has 5 electrons in its 2p orbital which are 1 less than required for ideal inert gas configuration, which is responsible for its high nuclear charge and smaller size than that of neon.
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